Double shue and moduli spaces of curves in genus 0

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Double shue and moduli spaces of curves in genus 0

() Double shue 1 / 26

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Introduction

Integral representation of the double shue relations Series representation of the stue relations Integral representation of the shue relations The stue relations in terms of integrals

Moduli spaces of curves ; double shue and forgetful maps Shue and moduli spaces of curves

Stue and moduli spaces of curves

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Introduction

For a p-tuple k= (k₁, . . . ,k_p)of positive integers and k₁>2, the multiple zeta valueζ(k)is dened as

ζ(k) = X

n1>...>np>0

1 n^k₁¹· · ·n^kp^p

.

These values satisfy two families of algebraic (quadratic) relations known as double shue, or shue and stue described below.

The stue comes from the above representation,

the shue comes from the integral representation of the MZV.

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Integral representation of the double shue relations Series representation of the stue relations

Combinatorial stue

The stue product of a p-tuple k= (k₀,k_p)(k₀= (k₁, . . . ,k_p−1)) and a q-tuple l= (l₀,l_q)(l₀= (l₁, . . . ,l_q−1)) is dened recursively by the formula:

(k)∗(l) = (k∗l₀)·l_q+ (k₀∗l)·k_p+ (k₀∗l₀)·(k_p+l_q) (1) and k∗() = ()∗k=k.

Ifσis a term of the formal sum k∗l, we will writeσ∈st(k,l).

Example

(n)∗(m) = (n,m) + (m,n) + (n+m)

(u)∗(v,w) = (u,v,w,) + (v,u,w) + (v,w,u) + (u+v,w) + (v,u+w)

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Stue and multiple zeta values

Proposition (Stue relations)

Let k= (k1, . . . ,kp)and l= (l1, . . . ,lq)be as above with k1, l1>2. Then we have:

ζ(k)ζ(l) =



 X

n1>...>np>0

1 n^k₁¹· · ·n^kp^p







 X

m1>...>mq>0

1 m^l₁¹· · ·m^lq^q





= X

σ∈st(k,l)

ζ(σ).

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Stue and multiple zeta values

Example

ζ(k)ζ(l) =

+∞

X

n=1 +∞

X

m=1

1 n^km^l

= X

n>m>0

1

n^km^l + X

m>n>0

1

m^ln^k +X

n=m

1 n^k⁺^l

=ζ(k,l) +ζ(l,k) +ζ(k+l).

General case.

We split the summation domain of the productζ(k)ζ(l) {0<n1< . . . <np} × {0<m1< . . . <mq}

into all the domains that preserve the respective order of the ni and the mj and the boundary domains where some ni are equal to some mj.

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Integral representation of the double shue relations shue and MZV

Combinatorial shue

Let(e₁,e)and(f₁,f)be respectively an n-tuple and an m-tuple of symbols. The shue product(e₁,e)_X(f₁,f)is dened recursively by:

(e₁,e)_X(f₁,f) =e₁·(e X(f₁,f)) +f₁·((e₁,e)_Xf) (2) and e X() = ()_Xe=e.

Let e and f be an n-tuple and an m-tuple, ifσis a term of the formal sum e Xf , we will writeσ∈k Xl.

Such an element is dened by a permutation of[[1,n+m]]that we shall denote by σ∈sh(k,l).

Example

XY XAB =XYAB+XAYB+XABY+AXYB+AXBY+ABXY

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Integral representation of the MZV

To the tuple k, with n=k1+· · ·+kp, we associate the n-tuple k= ( 0, . . . ,0

| {z }

k1−1times

,1, . . . , 0, . . . ,0

| {z }

kp−1times

,1) = (ε_n, . . . , ε₁)

and the dierential form:

ω_k=ω_k= (−1)^p dt₁

t1−ε₁ ∧ · · · ∧ dt_n tn−ε_n.

Then, setting∆_n={0<t1< . . . <tn<1}, direct integration yields:

ζ(k) = Z

∆_n

ω_k.

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Integral representation of the shue relations

Proposition (Shue relations)

Let k= (k1, . . . ,kp)and l= (l1, . . . ,lq)with k1, l1>2. Then Z

∆_n

ω_k Z

∆_m

ω_l = X

σ∈sh(k,l)

Z

∆_n+m

ωσ. (3)

Proof.

Let n=k₁+...+k_p and m=l₁+...+l_q. Then we have:

Z

∆_n

ω_k Z

∆_m

ω_l= Z

∆

dt1

1−t₁· · ·dtn

t_n

dtn+1

1−t_n+1· · ·dtn+m

t_n+m .

The set∆ ={0<t1< . . . <tn<1} × {0<tn+1< . . . <tn+m<1}can be up, to codimension 1 sets, split into a union of simplexes

∆ = a

σ∈sh([[1,n]],[[n+1,m]])

∆_σ with∆_σ={0<tσ(1)<tσ(2)< ... <tσ(n+m)<1}.

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Integral representation of the double shue relations The stue relations in terms of integrals

MZV and Integration over a square

Example of cubical representation of the MZVs

We haveζ(2) =R

∆₂dt2

t2

dt1

1−t1. The change of variables t2=x1and t1=x1x2gives:

ζ(2) = Z

[0,1]2

dx1

x₁

x1dx2

1−x₁x₂ = Z

[0,1]2

dx1dx2

1−x₁x₂.

This change of variables is nothing but the blow up of the point(0,0)in the projective plane, in n dimensions it corresponds to a sequence of blow up given by:

tn=x1, tn−1=x1x2, . . . , t1=x1...xn. (4)

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Example of cubical representation of the MZVs

Using the former change of variable :

t_n=x₁, t_n−1=x₁x₂, . . . , t₁=x₁...x_n. for n=4 we write the multiple zeta values as follows:

zeta(4) = Z

[0,1]⁴

d⁴x

1−x₁x₂x₃x₄ ζ(2,2) = Z

[0,1]⁴

x₁x₂d⁴x

(1−x₁x₂)(1−x₁x₂x₃x₄) ζ(2)ζ(2) =

Z

[0,1]⁴

1 1−x1x2

1

1−x3x4d⁴x. For any variablesαandβ we have the equality:

1

(1−α)(1−β) = α

(1−α)(1−αβ)+ β

(1−β)(1−βα)+ 1

1−αβ. (5)

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Example of cubical representation of the MZVs

Settingα=x1x2 andβ =x3x4and applying (5), we recover the stue relation:

ζ(2)ζ(2) = Z

[0,1]4

x1x2

(1−x₁x₂)(1−x₁x₂x₃x₄)+ x3x4

(1−x₃x₄)(1−x₃x₄x₁x₂)

+ 1

1−x₁x₂x₃x₄

d⁴x (6)

ζ(2)ζ(2) =ζ(2,2) +ζ(2,2) +ζ(4).

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General Case

Let k= (k₁, . . . ,k_p)be as above and n=k₁+· · ·+k_p.

We dene f_k₁,...,kp to be the function of n variables on[0,1]ⁿ given by:

f_k₁,...,kp(x₁, . . . ,x_n) = 1 1−x1· · ·xk1

x1· · ·xk1

1−x₁· · ·x_k₁x_k₁+1· · ·x_k₁+k2

x1· · ·xk1+k2

1−x₁· · ·x_k₁+k2+k3

· · · x1· · ·xk1+...+kp−1

1−x1· · ·xk1+···+kp

. (7)

Proposition

For all p-tuples of integers(k₁, . . . ,k_p)with k₁>2, we have (with n=k₁+· · ·+k_p):

ζ(k₁, . . . ,k_p) = Z

[0,1]ⁿ

f_k₁,...,kp(x₁, . . . ,x_n)dⁿx. (8)

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Notation ...

To derive the stue relations in general using integrals and the functions fk1,...,kp, we will use the following notation. Let k be a sequence(k1, . . . ,kp),

n=k1+· · ·+kp. We have n variables x1, . . . ,xn.

Notation

a= (a1, . . . ,ar), we will write ^Qa=a1· · ·ar. x= (x₁, . . . ,x_n)and x(k,1) = (x₁, . . . ,x_k₁)and

x(k,i) = (x_k₁+···+ki−1+1, . . . ,x_k₁+···+ki), so the x is the concatenation of sequences x(k,1)· · ·x(k,p). x(k,6i) = (x₁, . . . ,x_k₁+···+ki). If k= (k₀,k_p),

x₀=x(k,6p−1) = (x₁, . . . ,x_k₁+···+kp−1).

If l is a q-tuple with l1+· · ·+lq=m andσ∈st(k,l)then :

yσ is the sequence where x(k,i)(resp. x⁰(l,j)) is in the position of ki (resp.

lj) inσ.

Components ofσof the form ki+lj give rise to sequence x(k,i)x⁰(l,j) We remark that for eachσ∈st(k,l),^Qyσ=^Qx^Qx⁰.

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Integral Stue

Remark

Let(k₁, . . . ,k_p) = (k₀,k_p)be a sequence of integers. Then:

f_k₁,...,kp(x(k,1), . . . ,x(k,p)) =f_k₁,...,kp−1(x(k,6p−1))

Qx(k,6p−1) 1−^Qx(k,6p) (9)

=fk1,...,kp−1(x0) Qx₀

1−^Qx. (10)

Proposition

Let k= (k1, . . . ,kp)and l= (l1, . . . ,lq)be two sequences with n=k1+· · ·+kp

and m=l1+· · ·+lq. Then:

f_k₁,...,kp(x(k,1), . . . ,x(k,p))·f_l₁,...,lq(x⁰(l,1), . . . ,x⁰(l,q)) = X

σ∈st(k,l)

fσ(yσ). (11)

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Proof

We proceed by induction on the depth of the sequence. The recursion formula for the stue is given in (1).

If p=q=1: This is nothing but formula (5):

f_n(x(k,1))f_m(x⁰(l,1)) = 1

1−^Qx(k,61)· 1

1−^Qx⁰(l,61) = 1

1−^Qx · 1 1−^Qx⁰

fn(x(k,1))fm(x⁰(l,1))⁽=⁵⁾

Qx

(1−^Qx)(1−^Qx^Qx⁰)+

Qx⁰

(1−^Qx⁰)(1−^Qx⁰^Qx)+ 1

1−^Qx^Qx⁰. (12)

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Proof

Inductive step: Let(k₁, . . . ,k_p) = (k₀,k_p)et(l₁, . . . ,l_q) = (l₀,l_q)be two sequences.

f_k₀,kp(x₀,x(k,p))f_l₀,lq(x₀⁰,x⁰(l,q)) =f_k₀(x₀)

Qx(k,6p−1) 1−^Qx(k,6p) f_l₀(x₀⁰)

Qx⁰(l,6q−1) 1−^Qx⁰(l,6q)

(5)

=f_k₀(x₀)f_l₀(x₀⁰)·(^Qx(k,6p−1)^Qx⁰(l,6q−1)) Qx(k,6p)

(1−^Qx(k,6p))(1−^Qx(k,6p)^Qx⁰(l,6q))

+

Qx⁰(l,6q)

(1−^Qx⁰(l,6q))(1−^Qx⁰(l,6q)^Qx(k,6p))

+ 1

(1−^Qx(k,6p)^Qx⁰(l,6q))

.

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Proof

Expanding and using the remark we obtain:

f_k₀,kp(x₀,x(k,p))f_l₀,lq(x₀⁰,x⁰(l,q)) = f_k₀,kp(x)f_l₀(x₀⁰)

·

Qx^Qx⁰₀ 1−^Qx^Qx⁰

+ f_k₀(x₀)f_l₀,lq(x⁰)

·

Qx⁰^Qx₀

1−^Qx⁰^Qx + (f_k₀(x₀)f_l₀(x₀⁰))·

Qx₀^Qx⁰₀ 1−^Qx^Qx⁰. Hence, the product of functions f_k₁,...,kp and f_l₁,...,lq satises a recursion formula identical to the formula (1) that denes the stue product. Using induction, the proposition follows.

Corollary (integral representation of the stue)

Integrating the statement of the previous proposition over the cube and permuting the variables in each term of the LHS, we obtain:

ζ(k)ζ(l) = Z

[0,1]ⁿ

f_kdⁿxZ

[0,1]^m

f_ld^mx= Z

[0,1]^n+m

X

σ∈st(k,l)

fσdⁿ⁺^mx= X

σ∈st(k,l)

ζ(σ).

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Shue and moduli spaces of curves

Let k and l be as in the previous section, let n=k1+· · ·+kp and m=l1+· · ·+lq.

We will identify a point ofM₀,j+3with a sequence(0,z1, . . . ,zj,1,∞), the zi ∈P¹ all distinct and distinc from 0,1 and∞.

Φ_j is the open cell inM₀,j+3(R)which is mapped onto∆_j, the standard simplex, by the map: M₀,j+3→(P¹)^j

(0,z₁, . . . ,z_j,1,∞)7→(z₁, . . . ,z_j).

ω_k is the Kontsevich form corresponding to the tuple(k1, . . . ,kp) Then we have:

ζ(k₁, . . . ,k_p) = Z

Φ_n

ω_k.

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Proposition

Letβ be the map dened by:

M₀,n+m+3 β

−−−−−→ M₀,n+3× M₀,m+3

(0,z1, . . . ,zn+m,1,∞) 7−→ (0,z1, . . . ,zn,1,∞)×(0,zn+1, . . . ,zn+m,1,∞).

Then, if we write ti for the coordinate such that ti(0,z1, . . . ,zn+m,1,∞) =zi, we have:

β^∗(ω_k∧ω_l) = dt1

1−t₁ ∧ · · · ∧dtn

t_n ∧ dtn+1

1−t_n+1 ∧ · · · ∧dtn+m

t_n+m .

Furthermore, if forσ∈sh((1, . . . ,n),(n+1, . . . ,n+m))we writeΦ^σ_n₊_m orΦσ for the open cell ofM₀,n+m+3(R)in which the points are in the same order as their indices are inσ, we have:

β⁻¹(Φ_n×Φ_m) = a

σ∈sh((1,...,n),(n+1,...,n+m))

Φ^σ_n₊_m.

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Proof.

The rst part is obvious.

In order to show thatβ⁻¹(Φ_n×Φ_m) =`

Φ^σ_n₊_m we have to remember that a cell inM₀,n+m+3(R)is given by a cyclic order on the marked points. Let

X = (0,z1, . . . ,zn+m,1,∞)be a point inM₀,n+m+3(R)such that β(X)∈Φ_n×Φ_m. The values of the zi have to be such that:

0<z₁< . . . <z_n<1(<∞) and 0<z_n+1< . . . <z_n+m <1(<∞). (13) However there is no order condition relating z₁to z_n+1 for example. So, points on M₀,n+m+3(R)which are inβ⁻¹(Φ_n×Φ_m)are such that the z_i are compatible with (13). That is exactly a

σ∈sh((1,...,n),(n+1,...,n+m))

Φ^σ_n₊_m.

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Proposition

The shue product can be seen as the change of variables:

Z

Φ_n×Φ_m

ω_k∧ω_l= Z

β⁻¹(Φ_n×Φ_m)

β^∗(ω_k∧ω_l).

Proof.

Using the previous proposition, the right hand side of this equality is equal to X

σ∈sh((1,...,n),(n+1,...,n+m))

Z

Φ^σ_n+m

dt₁

1−t1 ∧ · · · ∧dt_n+m

tn+m .

Then we permute the variables and change their names in order to have an integral overΦ_n+m for each term.

As the form ₁^dt₋^σ(1)_t_σ(1) ∧ · · · ∧^dt_t^σ(n+m)

σ(n+m) does not have any pole on the boundary ofΦ^σ_n₊_m all the integrals are convergent.

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Moduli spaces of curves ; double shue and forgetful maps Stue and moduli spaces of curves

Stue and Moduli space of curves

The cubical coordinates onM₀,r are dened by u1=tr and ui =tr−i+1/tr−i+2

for i <r. This cubical system is well adapted to express the stue relations on the moduli spaces of curves.

Proposition

Letδbe the map dened by:

M₀,n+m+3 δ

−−→ M₀,n+3× M₀,m+3

(0,z₁, . . . ,z_n+m,1,∞) 7−→ (0,z_m+1, . . . ,z_m+n,1,∞)×(0,z₁, . . . ,z_m,z_m+1,∞).

Then, settingω_k=f_k(u₁, . . . ,u_n)dⁿu andω_l=f_l(u_n+1, . . . ,u_n+m)d^mu where the f_kare as in the work onRⁿ and the u_i are the cubical coordinates on moduli spaces, we have:

δ^∗(ω_k∧ω_l) =fk1,...,kp(u1, . . . ,un)fl1,...,lq(un+1, . . . ,un+m)dⁿ⁺^mu and

δ⁻¹(Φ_n×Φ_m) = Φ_n+m.

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Proof.

To prove the second statement, let X = (0,z1, . . . ,zn+m,1,∞)such that δ(X)∈Φ_n×Φ_m. Then the values of the zi are real and have to verify:

0<z1< . . . <zm<zm+1(<∞) and 0<zm+1< . . . <zn+m<1(<∞).

These conditions show that 0<z₁< . . . <z_m <z_m+1< . . . <1<∞, so (14) X ∈Φ_n₊_m.

To prove the rst statement, we claim thatδis expressed in cubical coordinates by:

(u₁, . . . ,u_n+m)7−→(u₁, . . . ,u_n)×(u_n+1, . . . ,u_n+m).

It is obvious to see that on the left hand factor the coordinates are not changed.

For the right hand factor we have to rewrite the expression of the right side in terms of the standard set of representative onM₀,m+3. We have:

(0,z1, . . . ,zm,zm+1,∞) = (0,z1/zm+1, . . . ,zm/zm+1,1,∞) = (0,t1, . . . ,tm,1,∞) in simplicial coordinates. This point is given in cubical coordinates onM₀,m+3by:

(tm,tm−1/tm, . . . ,t1/t2) = (zm/zm+1, . . . ,z1/z2) = (un+1, . . . ,un+m).

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As a consequence of this discussion and the results of section 3, we have the following proposition.

Proposition

Using the Cartier decomposition (11), the stue product can be viewed as the change of variables:

Z

Φ_n×Φ_m

ω_k∧ω_l= Z

δ⁻¹(Φ_n×Φ_m)

δ^∗(ω_k∧ω_l).

Remark

We should point out here the fact that the Cartier decomposition "does not lie in the moduli spaces of curves":

Forms appear in the decomposition which are not holomorphic on the moduli space.

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Remark

For example, in the Cartier decomposition of f₂,1(u₁,u₂,u₃)f₂,1(u₄,u₅,u₆), there appears the term:

u₁u₂u₄u₅du₁du₂du₃du₄du₅du₆ (1−u1u2u4u5)(1−u1u2u3u4u5u6) which is not a holomorphic dierential form onM₀,6.

However, it is a well-dened convergent form on the standard cell where it is integrated. Changing the numbering of the variables (which is allowed because the standard cell is symmetric under the permutation of the variable) gives the equality withζ(4,2).

This example represents the situation in the general case: when simply dealing with integrals, the non-holomorphic forms are not a problem.

However, in the context of framed motives they are.

Double shue and moduli spaces of curves in genus 0